Design of a Helical Compression Spring for Cyclic Loading Problem Design a compression spring for a dynamic load over a given deflection. Given The spring must give a minimum force of 60 lb and a maximum force of 150 lb over a dynamic deflection of 1.00 in. The forcing frequency is 1 000 rpm.

Question

Design of a Helical Compression Spring for Cyclic Loading

Problem Design a compression spring for a dynamic load over a given deflection.

Given The spring must give a minimum force of 60 lb and a maximum force of 150 lb over a dynamic deflection of 1.00 in. The forcing frequency is 1 000 rpm. A 10-year life of 1-shift operation is desired.

Assumptions Music wire (ASTM A228) will be used, since the loads are dynamic. Peening will be used to obtain a higher endurance strength.

Accepted Answer

See Figures 14-18, 14-19 and Table 14-11.

1 Find the number of cycles that the spring will see over its design life.

[latex] N_{ ext {life }}=1000 \frac{ rev }{\min }\left\lgroup\frac{60 min }{ hr } ight group \left\lgroup\frac{2080 hr }{ shift - yr } ight group (10 yr )=1.2 E 9 ext { cycles } [/latex] (a)

This large a number requires that an endurance limit for infinite life be used.

2 Find the mean and alternating forces from equation 14.16a:

[latex] \begin{array}{l} F_a=\frac{F_{\max }-F_{\min }}{2} \ F_m=\frac{F_{\max }+F_{\min }}{2} \end{array} [/latex] (14.16a)

[latex] \begin{array}{c} F_a=\frac{F_{\max }-F_{\min }}{2}=\frac{150-60}{2}=45 lb \ F_m=\frac{F_{\max }+F_{\min }}{2}=\frac{150+60}{2}=105 lb \end{array} [/latex] (b)

3 Assume a 0.207-in wire diameter from the available sizes in Table 14-2 (p. 791). Assume a spring index of 9 and calculate the mean coil diameter [latex]D[/latex] from equation 14.5 (p. 797).

[latex] C=\frac{D}{d} [/latex] (14.5)

[latex] D=C d=9(0.207)=1.863 ext { in } [/latex] (c)

4 Find the direct shear factor [latex]K_{s}[/latex] and use it to calculate the stress [latex] au _{i}[/latex] at the initial deflection (lowest defined force), and the mean stress [latex] au _{m}[/latex]:

[latex] K_s=1+\frac{0.5}{C}=1+\frac{0.5}{9}=1.056 [/latex] (d)

[latex] au_i=K_s \frac{8 F_i D}{\pi d^3}=1.056 \frac{8(60)(1.863)}{\pi(0.207)^3}=33875 psi [/latex] (e)

[latex] au_m=K_s \frac{8 F_m D}{\pi d^3}=1.056 \frac{8(105)(1.863)}{\pi(0.207)^3}=59281 psi [/latex] (f)

5 Find the Wahl factor [latex]K_{w}[/latex] and use it to calculate the alternating shear stress [latex] au _{a}[/latex] in the coil.

[latex] K_w=\frac{4 C-1}{4 C-4}+\frac{0.615}{C}=\frac{4(9)-1}{4(9)-4}+\frac{0.615}{9}=1.162 [/latex] (g)

[latex] au_a=K_w \frac{8 F_a D}{\pi d^3}=1.162 \frac{8(45)(1.863)}{\pi(0.207)^3}=27970 psi [/latex] (h)

6 Find the ultimate tensile strength of this music-wire material from equation 14.3 and Table 14-4 (p. 792) and use it to find the ultimate shear strength from equation 14.4 and the torsional yield strength from Table 14-8 (p. 803), assuming that the set has been removed and using the low end of the recommended range.

[latex] S_{u t} \cong A d^b [/latex] (14.3)

[latex] S_{u s} \cong 0.67 S_{u t} [/latex] (14.4)

[latex] \begin{array}{l} S_{u t}=A d^b=184649(0.207)^{-0.1625}=238507 psi \ S_{u s}=0.67 S_{u t}=159800 psi \end{array} [/latex] (i)

[latex] S_{y s}=0.60 S_{u t}=0.60(238507)=143104 psi [/latex] (j)

7 Find the wire endurance limit for peened springs in repeated loading from equation 14.13 (p. 804) and convert it to a fully reversed endurance strength with equation 14.18b (p. 814).

[latex] \begin{array}{l} S_{e w^{\prime}} \cong 45.0 kpsi (310 MPa ) \quad ext {for unpeened springs} \ S_{e w^{\prime}} \cong 67.5 kpsi (465 MPa ) \quad ext {for peened springs} \end{array} [/latex] (14.13)

[latex] S_{e s}=0.5 \frac{S_{e w} S_{u s}}{S_{u s}-0.5 S_{e w}} [/latex] (14.18b)

[latex] S_{e w}=67500 psi [/latex] (k)

[latex] S_{e s}=0.5 \frac{S_{e w} S_{u s}}{S_{u s}-0.5 S_{e w}}=0.5 \frac{67500(159800)}{159800-0.5(67500)}=42787 psi [/latex] (l)

8 The safety factor is calculated from equation 14.18a (p. 814).

[latex] N_{f_s}=\frac{S_{e s}\left(S_{u s}- au_i ight)}{S_{e s}\left( au_m- au_i ight)+S_{u s} au_a} [/latex] (14.18a)

[latex] \begin{aligned} N_{f s} &=\frac{S_{e s}\left(S_{u s}- au_i ight)}{S_{e s}\left( au_m- au_i ight)+S_{u s} au_a} \ &=\frac{42787(159800-33875)}{42787(59281-33875)+159800(27970)}=1.0 \end{aligned} [/latex] (m)

This is obviously not an acceptable design. To get some idea of what to change to improve it, the model was solved for a list of values of the spring index [latex]C[/latex] from 4 to 14, keeping all other parameters as defined above. The resulting values of coil diameter, free length, spring weight, and torsional fatigue safety factor are plotted in Figure 14-18. Note that the wire diameter was held constant to develop the variation of parameters with spring index as shown in Figure 14-18. If another parameter such as mean coil diameter [latex]D[/latex] were held constant instead, a different set of functions would result for the free length, weight, safety factor, etc.

The safety factor increases with decreasing spring index, so a reduction in our assumed value for [latex]C[/latex] will improve the safety factor even with no change in wire diameter. Note, however, that the free length increases exponentially with decreasing spring index. If package size is limited, we may not want to decrease the spring index too much in order to avoid excessive spring length. The coil diameter increases linearly with the spring index for a constant wire diameter. Spring weight decreases slowly with increasing spring index.

If we decrease the spring index from 9 to 7, keeping all other parameters the same, we will obtain an acceptable design in this case with [latex]N_{f}[/latex] = 1.3. Table 14-11 shows the results of this new calculation and of the calculations outlined below needed to complete the design. Figure 14-19 shows the modified-Goodman diagram for the final design. A summary of the changed values is

[latex] \begin{array}{llll} C=7 & D=1.45 in & K_w=1.21 & K_s=1.07 \ au_i=26743 psi & au_a=22705 psi & au_m=46800 psi & N_{f_s}=1.3 \end{array} [/latex] (n)

9 The spring rate is defined from the two specified forces at their relative deflection.

[latex] k=\frac{\Delta F}{y}=\frac{150-60}{1.0}=90 lb / ext { in } [/latex] (o)

10 To get the defined spring rate, the number of active coils must satisfy equation 14.7 (p. 797):

[latex] k=\frac{F}{y}=\frac{d^4 G}{8 D^3 N_a} [/latex] (14.7)

[latex] k=\frac{d^4 G}{8 D^3 N_a} \quad ext { or } \quad N_a=\frac{d^4 G}{8 D^3 k}=\frac{(0.207)^4 11.5 E 6}{8(1.45)^3(90)}=9.64 \cong 9 \frac{3}{4} [/latex] (p)

Note that we round it to the nearest 1/4 coil, as the manufacturing tolerance cannot achieve better than that accuracy. This makes the spring rate [latex]k[/latex] = 89 lb/in.

11 Assume squared and ground ends, making the total number of coils from Figure 14-9 (p. 796):

[latex] N_t=N_a+2=9.75+2=11.75 [/latex] (q)

12 The shut height can now be determined.

[latex] L_s=d N_t=0.207(11.75)=2.43 ext { in } [/latex] (r)

13 The initial deflection to reach the smaller of the two specified loads is

[latex] y_{ ext {initial }}=\frac{F_{ ext {initial }}}{k}=\frac{60}{89}=0.674 ext { in } [/latex] (s)

14 Assume a clash allowance of 15% of the working deflection:

[latex] y_{ ext {clash }}=0.15 y=0.15(1.0)=0.15 ext { in } [/latex] (t)

15 The free length (see Figure 14-8, p. 796) can now be found from

[latex] L_f=L_s+y_{ ext {clash }}+y_{ ext {working }}+y_{ ext {initial }}=1.82+0.15+1.0+0.674=4.25 ext { in } [/latex] (u)

16 The deflection to the shut height is

[latex] y_{s h u t}=L_f-L_s=4.25-2.43=1.82 ext { in } [/latex] (v)

17 The force at this shut-height deflection is

[latex] F_{ ext {shut }}=k y_{ ext {shut }}=89(1.82)=162 lb [/latex] (w)

18 The shut-height stress and safety factor are

[latex] au_{ ext {shut }}=K_s \frac{8 F_{ ext {shut }} D}{\pi d^3}=1.07 \frac{8(164)(1.45)}{\pi(0.207)^3}=72875 psi [/latex] (x)

[latex] N_{s_{ ext {shut }}}=\frac{S_{y s}}{ au_{ ext {shut }}}=\frac{143104 psi }{72875 psi }=2.0 [/latex] (y)

which is acceptable.

19 To check for buckling, two ratios need to be calculated, [latex]L_{f} / D[/latex] and [latex]y_{max} / L_{f}[/latex].

[latex] \begin{array}{c} \frac{L_f}{D}=\frac{4.25}{1.45}=2.93 \ \frac{y_{\max }}{L_f}=\frac{y_{ ext {initial }}+y_{ ext {working }}}{L_f}=\frac{0.674+1.0}{4.25}=0.39 \end{array} [/latex] (z)

Take these two values to Figure 14-14 (p. 801) and find that their coordinates are safely within the zones that are stable against buckling for either end-condition case.

20 The weight of the spring’s active coils from equation 14.12b (p. 802) is

[latex] W_a=\frac{\pi^2 d^2 D N_a \gamma}{4} [/latex] (14.12b)

[latex] W_a=\frac{\pi^2 d^2 D N_a \gamma}{4}=\frac{\pi^2(0.207)^2(1.45)(9.75)(0.285)}{4}=0.426 lb [/latex] (aa)

21 The natural frequency of this spring is found from equation 14.12a (p. 802) and is

[latex] \omega_n=\pi \sqrt{\frac{k g}{W_a}} rad / sec \quad f_n=\frac{1}{2} \sqrt{\frac{k g}{W_a}} Hz [/latex] (14.12a)

[latex] f_n=\frac{1}{2} \sqrt{\frac{k g}{W}}=\frac{1}{2} \sqrt{\frac{89(386)}{0.426}}=142 Hz =8521 \frac{ ext { cycles }}{ ext { min }} [/latex] (ab)

The ratio between the natural frequency and the forcing frequency is

[latex] \frac{8520}{1000}=8.5 [/latex] (ac)

which is sufficiently high.

22 The design specification for this A228 wire spring is

[latex] d=0.207 ext { in } \quad D_o=1.66 ext { in } \quad N_t=11.75, sq \& g \quad L_f=4.25 [/latex] (ad)

23 Detailed results are shown in Table 14-11. The files EX14-04 are on the CD-ROM. There are also alternate approaches to the solution of this example shown in separate Mathcad and TKSolver files on the CD-ROM with a letter added to their name.

Table 14-11Example 14-4 Helical Compression Spring for Dynamic Loading
Input	Variable	Output	Unit	Comments
1 000	[latex] rpm [/latex]		rpm	excitation frequency
7	[latex] C [/latex]			trial spring index
0.207	[latex] d [/latex]		in	available wire diameter (List Function)
1	[latex] y [/latex]		in	deflection of spring
'music	[latex] matl [/latex]			one of 'music, 'oiltemp, 'hdrawn, etc.
'sqgrnd	[latex] end [/latex]			one of 'plain, 'pgrnd, 'square, 'sqgrnd
'peen	[latex] surface [/latex]			one of 'unpeen or 'peen
'set	[latex] setflag [/latex]			set for a set spring—'unset otherwise
150	[latex] Fmax [/latex]		lb	maximum applied force
60	[latex] Fmin [/latex]		lb	minimum applied force
	[latex] Falt [/latex]	45	lb	alternating force
	[latex] Fmean [/latex]	105	lb	mean force
	[latex] Fshut [/latex]	164	lb	force at shut height
	[latex] k [/latex]	89	lb/in	spring rate with [latex]N_{a}[/latex] rounded
	[latex] Na [/latex]	9.75		no. of active coils—rounded to 1/4 coil
	[latex] Nt [/latex]	11.75		no. of total coils
	[latex] D [/latex]	1.45	in	mean coil diameter
	[latex] Dout [/latex]	1.66	in	outside coil diameter
	[latex] Din [/latex]	1.24	in	inside coil diameter
	[latex] Ks [/latex]	1.07		static factor—direct shear—Eq. 13.8
	[latex] Kw [/latex]	1.21		Wahl Factor—Eq. 13.9
	[latex] tauinit [/latex]	26 743	psi	shear stress at installed length
	[latex] taushut [/latex]	72 875	psi	stress at shut height
	[latex] taualt [/latex]	22 705	psi	alternating shear stress for fatigue
	[latex] taumean [/latex]	46 800	psi	mean shear stress for fatigue
	[latex] Sut [/latex]	238 507	psi	tensile strength—Eq. 13.3 & Table 13-4
	[latex] Sus [/latex]	159 800	psi	ultimate shear strength—Eq. 13.4
	[latex] Sys [/latex]	143 104	psi	shear yield based on Table 13-6
	[latex] Sew [/latex]	67 500	psi	wire endurance limit—Eq. 13.12
	[latex] Ses [/latex]	42 787	psi	fully reversed endurance limit—Eq. 13.16b
	[latex] Nf [/latex]	1.3		safety factor—fatigue—Eq. 13.14
	[latex] Nshut [/latex]	2		safety factor—shut height (yielding)
	[latex] Lf [/latex]	4.25	in	free length
	[latex] Lshut [/latex]	2.43	in	shut height
	[latex] yinit [/latex]	0.67	in	initial deflection at assembly
	[latex] yshut [/latex]	1.82	in	deflection to shut height
	[latex] nf [/latex]	142	Hz	natural frequency in Hz
	[latex] FreqFac [/latex]	8.5		ratio of nat. freq to excitation freq.

Table 14-4 Coefficients and Exponents for Equation 14.3 Source: Reference 1
ASTM#	Material	Range		Exponent [latex]b[/latex]	Coefficient [latex]A[/latex]		Correlation Factor
ASTM#	Material	mm	in	Exponent [latex]b[/latex]	MPa	psi	Correlation Factor
A227	Cold drawn	0.5–16	0.020–0.625	–0.182 2	1 753.3	141 040	0.998
A228	Music wire	0.3–6	0.010–0.250	–0.1625	2 153.5	184 649	0.9997
A229	Oil tempered	0.5–16	0.020–0.625	–0.183 3	1 831.2	146 780	0.999
A232	Chrome-v	0.5–12	0.020–0.500	–0.145 3	1 909.9	173 128	0.998
A401	Chrome-s.	0.8–11	0.031–0.437	–0.093 4	2 059.2	220 779	0.991

Table 14-8 Maximum Torsional Yield Strength [latex]S_{ys}[/latex] for Helical Compression Springs in Static Applications Bending or Buckling Stresses Not Included. Source: Adapted from Ref. 1
Material	Maximum Percent of Ultimate Tensile Strength
Material	Before Set Removed (Use Eq. 13.10b)	After Set Removed (Use Eq. 13.9b)
Cold-drawn carbon steel (e.g., A227, A228)	45%	60–70%
Hardened and tempered carbon and low-alloy steel (e.g., A229, A230, A232, A401)	50	65–75
Austenitic stainless steel (e.g., A313)	35	55–65
Nonferrous alloys (e.g., B134, B159, B197)	35	55–65

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